# Property singular support of the convolution of distributions

Let $u \in \mathcal{D}'(\Omega)$ and $U \subset \Omega$ open. By definition we say that $u \in \mathcal{E}(U)$ if $\exists u(x) \in \mathcal{E}(U)$ such that \begin{align*} \displaystyle \langle \varphi , u \rangle = \int_U u(x) \varphi(x) dx , \forall \varphi \in \mathcal{D}(U) \end{align*} and singular support of $u \in \mathcal{D}'(\Omega)$ is the set of all points $x \in \Omega$ such that not exists an open neighborhood $U\subset \Omega$ wich $u_{|U}$ is regular function, i.e. \begin{align*} \displaystyle \mathrm{singsupp}(u) = \lbrace x \in \Omega : \nexists U \subset \Omega : u_{|U} \in \mathcal{E}(U) \rbrace \end{align*} and then if $U_{max}$ is maximum open of $\Omega$ with the distribution $u \in \mathcal{E}(U_{\max})$ then \begin{align*} \displaystyle \mathrm{singsupp}(u) = \Omega \setminus U_{max} \end{align*} I'm trying to make a little better the proof that if $u$ or $v$ is a distrubution with compact support then

\begin{align*} \mathrm{singsupp}(u \ast v) \subset \mathrm{singsupp}(u) + \mathrm{singsupp}(v) \end{align*}

this proof it can be found in "Distribution Theory" by Duistermaat, page 130. It uses a regular version of the lemma Urysohn (Lemma 2.19 page 29), or as here Proof of regular version of the Urysohn lemma

As mentioned, I'm trying to better express this demonstration (and this proof is slightly different).

Proof. We place $L=\mathrm{singsupp}(u)$ and $M=\mathrm{singsupp}(v)$. Consider $\mathrm{Int}(L)\subset L$ and $\mathrm{Int}(M) \subset M$, by Urysohn lemma $\forall \delta > 0$ small \begin{align*} \displaystyle \exists \psi \in \mathcal{D}(\mathbb{R}^n) : \mathrm{supp}(\psi) \subset L(\delta) \subset \mathrm{Int}(L) \end{align*} with $\psi(x)=1$ $\forall x \in L(\delta)$ and $0 \leq \psi(x) \leq 1$ $\forall x \in \mathrm{Int}(L)$, where $L(\delta)$ is a compact. Similary \begin{align*} \displaystyle \exists \eta \in \mathcal{D}(\mathbb{R}^n) : \mathrm{supp}(\eta) \subset M(\delta) \subset \mathrm{Int}(M) \end{align*} with $\eta(x)=1$ $\forall x \in M(\delta)$ and $0 \leq \eta(x) \leq 1$ $\forall x \in \mathrm{Int}(M)$, where $M(\delta)$ is also a compact set. We define $u_1=\psi u$ and $u_2=(1-\psi)u$, and also $v_1=\eta v$ and $v_2=(1-\eta)v$, then $u=u_1+u_2$ and $v=v_1+v_2$, with $\mathrm{supp}(u_1) \subset \mathrm{supp}(\psi) \subset L(\delta)$, and $\mathrm{supp}(v_1) \subset \mathrm{supp}(\eta) \subset M(\delta)$, therefore $u_1, v_1 \in \mathcal{E}'(\mathbb{R}^n)$ are distribution with compact support. Moreover $u_2 , v_2 \in \mathcal{E}(\mathbb{R}^n)$ (WHY?). Consequently of the four terms in following convolution, only the last three are regular functions \begin{align*} \displaystyle u \ast v = u_1 \ast v_1 + u_1 \ast v_2 + u_2 \ast v_1 + u_2 \ast v_2 \end{align*} this implies that \begin{align*} \displaystyle \mathrm{singsupp}(u \ast v) &= \mathrm{singsupp}(u_1 \ast v_1) \\ &\subset \mathrm{supp}(u_1 \ast v_1) \\ &\subset \mathrm{supp}(u_1) + \mathrm{supp}(v_1) \\ &\subset L(\delta)+M(\delta) \\ &\subset L+M \end{align*}

• What happens if one of the distributions, e.g. $u$, has a non compact singular support. Why do you get $u_2\in\mathcal{E}$? Jul 5 '16 at 16:06
• In addition: you don't need that both $u_2$ and $v_2$ are smooth since the convolution of a distribution with a smooth function is a smooth function. Jul 5 '16 at 16:10
• Could you clarify what properties $L(\delta)$ has in addition to being compact? Jul 5 '16 at 16:16
• I assume that $u$ or $v$ is a distribution with compact support just so because $u \ast v$ is well defined, otherwise the convolution of two distributions is not well defined. Instead you need to have $u_2$ and $v_2$ as regular function, otherwise I can not deduce that $u_1 \ast v_2$, $u_2 \ast v_1$, $u_2 \ast v_2$ are regular functions. See link of the proof of regular version of Urysohn lemma. Jul 5 '16 at 16:17
• Of course, you are right that both $u_2$ and $v_2$ need to be smooth. But $u$ or $v$ might have noncompact singular support isn't it? Jul 5 '16 at 16:20

1) Since you have $L(\delta)\subseteq L$ and not $L\subseteq L(\delta)$, I do not see why $(1-\psi)u$ should be a smooth function.
2) One of your distributions $u$ or $v$ might have non-compact singular support. Assume $u$ is the one where $\operatorname{sing supp} u$ is non-compact. Since $\operatorname{supp}\psi$ is compact there is no chance that $1-\psi$ can vanish on the whole singular support.
I think that these two problems are the reason why in the "standard proof" (e.g. given in the book of Hörmander) only the compactly supported distribution is split in a smooth and a singular part. In addition usually it is assumed that $$\psi(x)=1 \text{ for } x\in K \text{ and } \operatorname{supp}\psi\subset B(K,\delta)$$ since then obviously $(1-\psi)u\in\mathcal{E}(\mathbb{R}^{n})$. On the other hand, in this case you have to show that also the convolution of the singular part with $v$ is smooth on a certain open set.
• According to your last observation, we have that $(1-\psi)u \in \mathcal{E}(\mathbb{R}^n)$ since it vanishes inside the compact $K$ ? note that for me, $M(\delta)=M' + \overline{B}(0, \delta)$ where essentially $\psi(x)=1$ $\forall x \in M'$ and $M'$ is another compact, so I have applied the Urysohn lemma for $\Omega=Int(M)$. Perhaps I failed to specify this. Jul 5 '16 at 18:19
• Yes, $(1-\psi)$ vanishes on $\operatorname{sing supp} u$ and hence $(1-\psi)u\in\mathcal{E}(\mathbb{R}^{n})$. In your argument $M(\delta)$ is a proper subset of $\operatorname{sing supp} u$, so there are points $x\in \operatorname{sing supp}u$ with $1-\psi(x)\neq 0$, so I don't see why $(1-\psi)u$ has to be smooth. Jul 5 '16 at 18:26
• I can do this, if I take $\mathrm{singsupp}(u)$ compact, but how do I deduce the same thing for $(1-\eta) v$ with $v \in \mathcal{D}'(\Omega)$ with not compact support? Jul 5 '16 at 18:44
• @JohnMartin: I think the above argument you need that $(1-\psi)$ is zero on the singular support of $u$ and $(1-\eta)$ is zero on the singular support of $v$. Could you add your argument for $(1-\psi)u\in\mathcal{E}(\mathbb{R}^{n})$ to your question since I still don't see it? Jul 6 '16 at 10:42